If the function is 
and f(4) − f(1) = f '(c)(4 − 1) then there is not any answer.
Given function is and f(4) − f(1) = f '(c)(4 − 1).
In this question we have to apply the mean value theorem, which says that given a secant line between points A and B, there is at least a point C that belongs to the curve and the derivative of that curve exists.
We begin by calculating f(2) and f(5):
f(2)=
f(2)=1
f(5)=
f(5)=1
And the slope of the function:
(x)=
(c)=0
Now,
=-2
=0
-2 is not equal to 0. So there is not any answer.
Hence if the function is and f(4) − f(1) = f '(c)(4 − 1) then there is not any answer.
Learn more about derivatives at brainly.com/question/23819325
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Answer:
Slope =3/8
y-intercept is:(0,6)
Step-by-step explanation:
Answer:
x = 129° ( linear pair then vertically opp angles)
The equation is <span>h(t) = -16t2 + 1600</span>.
When the package hits the ground, h(t) = 0: 0 = -16t2 + 1600 To solve by factoring: 0 = -16(t2-100)0 = -16(t+10)(t-10) t = -10 and +10 seconds.
Since we aren't interested in negative time, t = 10 seconds.
